\(\left\{{}\begin{matrix}2x^3=y+\dfrac{1}{y}\\2y^3=x+\dfrac{1}{x}\end{matrix}\right.\left(x,y\ne0\right)\)
\(\Leftrightarrow2\left(x^3-y^3\right)=y-x+\dfrac{1}{y}-\dfrac{1}{x}\)
\(\Leftrightarrow2\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)+\dfrac{x-y}{xy}=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+1+\dfrac{1}{xy}\right)=0\)
Xét \(x^2+y^2\ge2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)+2xy\ge2.2xy+2xy=6xy\)
\(\Leftrightarrow2x^2+2y^2+2xy+1+\dfrac{1}{xy}\ge6xy+\dfrac{1}{xy}+1\ge2\sqrt{6xy.\dfrac{1}{xy}}+1=2\sqrt{6}+1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\6xy=\dfrac{1}{xy}\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{\sqrt[4]{6}}}\)Suy ra hệ phương trình có nghiệm với mọi x=y và x,y khác 0.
\(\left\{{}\begin{matrix}2x^3=y+\dfrac{1}{y}\\2y^3=x+\dfrac{1}{x}\end{matrix}\right.\left(x,y\ne0\right)}\)
